A point charge of 10^{-8} text{ C} is placed | vedclass

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Question

(C) The work done by an external agent is equal to the change in potential energy of the system.$W_{	ext{ext}} = \Delta U = U_f - U_i$$W_{	ext{ext}}

Vedclass · Accepted Answer

$30 	imes 10^{-6}$

Vedclass · Answer

(C) The work done by an external agent is equal to the change in potential energy of the system.$W_{	ext{ext}} = \Delta U = U_f - U_i$$W_{	ext{ext}} = \frac{1}{4\pi\epsilon_0} q_1 q_2 \left( \frac{1}{r_f} - \frac{1}{r_i} ight)$Given:$q_1 = 10^{-8} 	ext{ C}$, $q_2 = 2 	imes 10^{-6} 	ext{ C}$$r_i = \sqrt{4^2 + 4^2 + 2^2} = \sqrt{16 + 16 + 4} = \sqrt{36} = 6 	ext{ m}$$r_f = \sqrt{2^2 + 2^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3 	ext{ m}$Substituting the values:$W_{	ext{ext}} = (9 	imes 10^9) 	imes (10^{-8} 	imes 2 	imes 10^{-6}) 	imes \left( \frac{1}{3} - \frac{1}{6} ight)$$W_{	ext{ext}} = (9 	imes 10^9) 	imes (2 	imes 10^{-14}) 	imes \left( \frac{2-1}{6} ight)$$W_{	ext{ext}} = 18 	imes 10^{-5} 	imes \frac{1}{6} = 3 	imes 10^{-5} 	ext{ J}$$W_{	ext{ext}} = 30 	imes 10^{-6} 	ext{ J}$

$A$ point charge of $10^{-8} \text{ C}$ is placed at the origin. The work done in moving a point charge of $2 \mu\text{C}$ from point $A(4, 4, 2) \text{ m}$ to $B(2, 2, 1) \text{ m}$ is . . . . . . $\text{J}$.

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